Pengertian KapasitorPengertian Kapasitor Dua penghantar (plat) berdekatan yang diberi
muatan sama tetapi berlawanan jenis disebut kapasitor.
Sifat menyimpan energi listrik / muatan listrik. Kapasitas suatu kapasitor (C) ditentukan oleh Kapasitas suatu kapasitor (C) ditentukan oleh
perbandingan antara besar muatan Q dari salah satu penghantarnya dengan beda potensial Vsatu penghantarnya dengan beda potensial Vantara kedua penghantar itu.
Kegunaan KapasitorU k hi d i j di l li ik d Untuk menghindari terjadinya loncatan listrik pada rangkaian2 yang mengandung kumparan bila tiba2 diputuskan arusnya.diputuskan arusnya.
Rangkaian yang dipakai untuk menghidupkan mesin mobil
Untuk memilih panjang gelombang yang ditangkap oleh pesawat penerima radio.
Bentuk kapasitor Kapasitor bentuk keping sejajar Kapasitor bentuk keping sejajar Kapasitor bentuk bola sepusat Kapasitor bentuk silinderKapasitor bentuk silinder
MacamMacam--Macam KapasitorMacam Kapasitor
Capacitor:Capacitor:Capacitor:Capacitor:A Device to Store EnergyStore Energy in the Form of an Electrican Electric Field
KapasitansiKapasitansi C adalah jumlah muatan yang disimpan dibagi dengan tegangandisimpan dibagi dengan tegangan
Q = CV
Satuan untuk kapasitansi
Q CV
Satuan untuk kapasitansi
Coulomb/Volt = FaradCoulomb/Volt Farad
Capacitanceh f hthe amount of charge
stored divided bythe voltage is thethe voltage is the capacitance C
unit of capacitanceQ = CV
unit of capacitance Coulomb/Volt = Farad
-
-
-
+
+
+
-
-
-
-
-
-
+
+
+
+
+
+
-
-
-
-
+
+
Kapasitor Pelat Paralel Dua pelat paralel yang luasDua pelat paralel yang luas luas A jarak antar pelat d jarak antar pelat d kerapatan muatan sama ttp berbeda tanda
C Q/Vd
C = Q/V= A/(d/0)d A/(d/0)
C = 0A/d 0Untuk menaikkan nilai kapasitansi
b k l
+ membesarkan luasanmemperkecil celah
Bila ukuran ke dua pelat cukup besardibanding jarak celah d Edibanding jarak celah d
Medan diantara ke dua pelat mempunyai nilai seragam
E = /E = /0Beda potensial diantarak d l V = Ed = d/0
+
ke dua pelat V Ed d/0C = Q/V = Q/(Q/4 R) = 4 RC = Q/V = Q/(Q/40 R) = 40 R
Kapasitansi bergantung pada bentuk geometri dariKapasitansi bergantung pada bentuk geometri dari permukaan.
Capacitance Depends onp p
Separation of the plates (d) [d in m]Separation of the plates (d) [d in m]
d1C d
Plate Area (A) [A in m2]AC
Dielectric Constant () [ in C2/N-m2]C
( ) [ / ]
Show Demo Model, calculate its capacitance , and showhow to charge it up with a battery.
Circular parallel plate capacitor
r = 10 cm
g p y
r r
s
A = r2 = (.1)2A = .03 m 2
S = 1 mm = .001 m
SAC 0=
S
001.03.)10( 11=C }Farad
VoltCoulomb
FC 10103 =
pFC 300= p = pico = 10-12
1. Consider two plates separated by d=1.5 cm , where the electric field between them is 100 V/m and the charge onelectric field between them is 100 V/m, and the charge on the plates is 30.0 C. What is the capacitance?
The Capacitance is:The Capacitance is:2.0 10-5 F Or 20 microfarad.
2. What is the capacitance of a capacitor if the charge is 0 075 C held at V = 400 V0.075 C held at V = 400 V
C = Q/V =0.075 C/400 V= 1.875 X 10-4 f = 187.5 F
Dielectrics K Capacitors are usually constructed with an
insulating material between the plates This helps prevent a breakdown or arcing This helps prevent a breakdown or arcing
between the plates at higher voltages The plates can be placed closer together withoutThe plates can be placed closer together without
danger of touching reducing d The capacitance is increased by a factor K
(dielectric constant) The formula for capacitance is now
C = K0 Ad0 d
Permitivitas Dielektrik = K0K adalah konstanta dielektrik
tergantung pada material dielektrik g g pumumnya nilai K = 1 sampai 10
Bila dielektrik sepenuhnya mengisi volume diantara pelat-pelat :
E berkurang oleh faktor K E = / = /K E berkurang oleh faktor K E = / = /K0 V berkurang oleh faktor K V = d/K V berkurang oleh faktor K V = d/K0 C meningkat oleh faktor K C = K A/d C meningkat oleh faktor K C = K0A/d
Material KVacuum 1.00000Gl 5 10Glass 5-10Mica 3-6Mylar 3.1P ffi W 2 1 2 5Paraffin Wax 2.1 2.5Plexiglas 3.40Polyethylene 2.25P l i l hl id 3 18Polyvinyl chloride 3.18
Teflon 2.1Porcelain 6.0 8.0
Germanium 16Rubber 2.5 3.0
Water 80 4Water 80.4Glycerin 42.5Liquid ammonia(-78C) 25Benzene 2 284Benzene 2.284Air(1 atm) 1.00059Air(100 atm) 1.0548
Question: The energy content of a charged capacitor resides in itsin its
(a) plates (b) potential difference(c) charge (d) electric field
Answer: d
Question: The plates of a parallel-plate capacitor of capacitance C are brought together to one-third their original separation. The capacitance is now
(a) C/9( )(b) C/3(c) 3C(d) 9C(d) 9C
Answer: c
Examplep Two rectangular sheets of copper foil 16 X 20 cm are
separated by a thin layer of paraffin wax 0 2 mm thickseparated by a thin layer of paraffin wax 0.2 mm thick. Calculate the capacitance if the dielectric constant for the wax is 2.4.
dAKC o=
m10X2mm2.0dm032.0cm320)cm20)(cm16(A
4
22
=====
pf3400f10X43Cm10X2
)m032.0)(4.2)(Nm/C10X854.8(C
9
4
22212
=
pf3400f10X4.3C ==
Energy Storage in CapacitorsEnergy Storage in Capacitors
The energy stored in a capacitor is just theThe energy stored in a capacitor is just the work necessary to separate the charges in the first place.p
As you keep moving charges, you have a greater force to overcome because the gpotential keeps increasing
The total work is just the total charge times j gthe average voltage or W = QV/2.
Energy Storage in CapacitorsTotal charge stored on capacitor
0
Q
W = V dq V = q / CTotal charge stored on capacitor
0
W ( /C)d Q2/2CQ1 Q
0W = (q/C)dq = Q2/2C
Energy = 12
QV
Q CVQVCVW 21
221 ==
Q =CVE U 1 CV 2 1 Q
2Energy =U =
2CV 2 =
2QC
where is the energy located?
W = C V2/2 = C (E d)2/2
= (0 A / d) E2 d2 / 2 ( 0 )= (0 E2/2)(A d)= (energy density)(volume) (energy density)(volume)
0E2/2 is the energy density0the energy is where the electric field is
between the plates
Misal kapasitor 1.2 F diberi muatan hingga berpotensial 3 kV, berapa energi yang tersimpan didalamnya?
QVCVW 21221 == W = 5.4 Joule
Hubungan paralel kapasitorcc
Parallel:V bernilai samauntuk ke dua kapasitor.Kapasitansi tinggal di jumlahkanjumlahkan.
Ceq = Qtotal/VCeq = Q1/V1 + Q2/V2V = V = VV1 = V2 = VCeq = C1 + C2
Hubungan Paralel Kapasitor
; ; ; ; 332211 VCQVCQVCQVCQ p====
321 CCCCp ++=
Kapasitor yang dihubungkan paralel, tegangan antara ujung2 kapasitor adalah sama, sebesar V.ujung kapas tor adalah sama, sebesar V.
Hubungan seri kapasitor
Seri : V tidak sama pada masing2Seri : V tidak sama pada masing2 kapasitor. Tetapi Q sama (mengapa?)
1/ Ceq = V1/ Q1 + V2/ Q2qQ1 = Q2 = Q1/ Ceq = 1/ C1 + 1/ C2
Hubungan seri kapasitor
adcdbcab CQV
CQV
CQV
CQV ==== ; ; ;
1111CCCC
++=sCCCC 321
Kapasitor yang dihubungkan seri akan mempunyai
321 CCCCs
Kapasitor yang dihubungkan seri akan mempunyai muatan yang sama.
321 QQQQ ===
Dua kapasitor dihubungkan seri yang dihubungkan dengan sumber 1000 V. Tentukan:
a. C-subsitusi (gabungan)b. Muatan masing-masing kapasitorc. Beda potensial pada ujung masing-masing
kapasitorkapasitor d. Energi yang tersimpan dalam susunan kapasitor
11111Penyelesaian:
pFpFCCCsub 61
31111
21
+=+=a. pFC sub 2=b. q1 = q2 q = Csub.V = (2 x 10-12 F) (1000 V) = 2 nC
c. V 667103102
12
9
1
11 ===
FxCx
CqV V 333
106102
12
9
2
22 ===
FxCx
CqV
Energi dalam d. g
( )( ) J 107.6V 66710221
21 79
111 === xCxVqW
( )( ) J 103.3V 33310221
21 79
222 === xCxVqW
22
( ) J x1010J 10 x 3.3 6.7 -77 =+= subW
Heart Muscle Cells The cell wall acts as a capacitor since charges are separated The cell wall acts as a capacitor since charges are separated
on the interior and exterior surfaces. The voltage is in the millivolt range.
Molecules in the cell wall are polarized. The charge on the outside of the wall is on the order of 10-8of the wall is on the order of 10 8Coulomb.
When the heart beats the wallWhen the heart beats, the wall depolarizes in a wave as shown in this picture. Here the wave proceeds from l f i h Af d l i i ileft to right. After depolarization is complete, the cell wall repolarizes.
This cycle gives rise to an electrical signal that can be detected externally with an EKG (elektrokardiogram) machine.
EKG TracingEKG Tracing
Th P i t ti f th h b t i ThThe P wave is contraction of the upper chambers or atria. The QRS is the ventricles. The wave proceeds from left to right and toward the front, then down to the left and toward the rear of the heart. The T wave is the re-polarization of the muscle in preparation for the next beat. Heart defects show up as variations in the wave pattern.in the wave pattern.